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We have the following DE: $$ \dfrac{dy}{dx} = \dfrac{x^2 + 3y^2}{2xy}$$

I don't know how to solve this. I know we need to write it as $y/x$ but I don't know how to in this case.

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Right-hand side is "homogeneous of order zero" type. Define $v=y/x$ and rewrite your equation in terms of $x$ and $v$. That is going to give a linear equation. –  Maesumi Feb 21 '13 at 21:17
    
@Maesumi Which is what I have trouble doing –  Ylyk Coitus Feb 21 '13 at 21:18
    
@Maesumi. Are you sure this method yields a linear equation? There will be a $v^{-1}$ term. –  Thomas E. Feb 21 '13 at 21:25
    
@ThomasE. Sorry! I meant to say it will be separable equation! –  Maesumi Feb 21 '13 at 21:27
    
"I like coitus"? Really? (If that's your actual name, I'm going to feel terrible about bringing this up, but I suspect you were just being whimsical.) –  Cameron Buie Feb 21 '13 at 21:42

2 Answers 2

up vote 1 down vote accepted

You are asked to solve, in terms of $\frac{y}{x}$, the following ODE:

$$\frac{\operatorname{d}\!y}{\operatorname{d}\!x} = \frac{x^2+3y^2}{2xy}$$

The way to do this is to make a substitution. Solving differential equations is like grown-up integration. Just as we used integration by substitution to solve ordinary integrals, we can do the same to solve ODEs.

The dependent variable in your ODE is $y$ and so we aim to replace it. Let us call $\frac{y}{x}=u$. We want to replace all of the $y$s, and so we use $y=ux$. As with integration by substitution, $u$ becomes our new independent variable. Let us see what we can do:

$$\frac{\operatorname{d}\!y}{\operatorname{d}\!x}= \frac{\operatorname{d}(ux)\!}{\operatorname{d}\!x}=x\frac{\operatorname{d}\!u}{\operatorname{d}\!x}+u$$

$$\frac{x^2+3y^2}{2xy} = \frac{x^2+3u^2x^2}{2x\,ux} = \frac{1+3u^2}{2u}$$

Putting these two substitutions together gives:

$$\frac{\operatorname{d}\!y}{\operatorname{d}\!x} = \frac{x^2+3y^2}{2xy} \implies x\frac{\operatorname{d}\!u}{\operatorname{d}\!x}+u=\frac{1+3u^2}{2u} \implies x\frac{\operatorname{d}\!u}{\operatorname{d}\!x}=\frac{1+u^2}{2u}$$

Taking the reciprocal of both sides we see that:

$$\frac{1}{x}\frac{\operatorname{d}\!x}{\operatorname{d}\!u}=\frac{2u}{1+u^2} \implies \int \frac{\operatorname{d}\!x}{x} = \int \frac{2u\operatorname{d}\!u}{1+u^2} \implies \ln|x|=\ln|k(1+u^2)|$$

It follows that $x=k(1+u^2)$ for some constant of integration. Given that $u=\frac{y}{x}$ we have:

$$x=k(1+u^2) \implies x=k\left(1+\frac{y^2}{x^2}\right) \implies y = \pm \frac{x}{k}\sqrt{k(x-k)}$$

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If you gave this answer earlier, I'd have awarded it since it is very thorough. I however already solved the ODE using Maesumi's answer before you gave this one, so I'd consider it unfair if I ended up giving it to you. I rated up your answer however, and I appreciate it a lot! –  Ylyk Coitus Feb 21 '13 at 22:31
    
@YlykCoitus That's absolutely fine. It's your question and your vote after all. I'll have to reply faster next time. –  Fly by Night Feb 21 '13 at 22:34

$y=vx$ so $y'=xv'+v$. Your equation is $xv'+v={{1 \over{2v}}+{{3v} \over {2}}}$. Now clean up to get $xv'={{v^2+1}\over{2v}}$. Now separate ${2v dv \over {v^2+1}} = {dx \over x}$.

Edit

${{x^2+3y^2} \over {2xy}} = {{{x^2}\over{2xy}}+{{3y^2}\over{2xy}}}={ x \over {2y}}+{{3y}\over{2x}}$

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I understand this, except for one thing: where did the x'es go at the RHS of the equation? Why isn't it $\dfrac{x}{2v} + \dfrac {3v}{2x}$ –  Ylyk Coitus Feb 21 '13 at 21:32
    
@YlykCoitus: $\frac{x^{2}}{2xy}=\frac{x}{2y}=\frac{1}{2v}$ and $\frac{3y^{2}}{2xy}=\frac{3y}{2x}=\frac{3v}{2}$. –  Thomas E. Feb 21 '13 at 21:35
    
@ThomasE. Oh, because you take the derivative? –  Ylyk Coitus Feb 21 '13 at 21:38
    
@YlykCoitus You separate the RHS to smaller fractions. –  Maesumi Feb 21 '13 at 21:39
    
@YlykCoitus I'm unsure how this actually solves the ODE. It certainly changes it into another one. –  Fly by Night Feb 21 '13 at 22:26

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